Solving Sampling Distributions in AP Statistics: Unbiased Estimators

[bjr_jyd15]

I need some help understanding sampling distributions. Here's the Q:

Prepare the sampling distribution x-bar of a random sample of n=2 drawn without replacement from a finite population of size n=5, whose elements are the numbers 3,5,7,9,11. Is the mean (mu x-bar) and unbiased estimator of mu?

So far I figured E(x-bar)=7. But I'm not sure how to determine the rest? Is there a formula I need for unbiased estimators?

I know there aren't too many taking AP stat out here, so ANY help would be great!

 

[Nate810]

The sampling distribution x-bar can be determined by calculating the mean of all possible samples of size two from the finite population of 5 numbers. Since there are 10 possible samples, you can calculate the sample means for each. For example, one possible sample is (3,5), which has a sample mean of 4. The next possible sample is (3,7), which has a sample mean of 5, and so on. Once you have calculated the sample means for all 10 possible samples, you can list them to create the sampling distribution. The mean (mu x-bar) is an unbiased estimator of mu if the expected value of the sampling distribution equals the true population mean (mu). In this case, the expected value of x-bar is 7, which is equal to the true population mean. So yes, the mean (mu x-bar) is an unbiased estimator of mu.

 

[wais]

Hi there,

Sampling distributions can be a tricky concept to grasp, but I will do my best to explain it to you.

First, let's define what a sampling distribution is. A sampling distribution is the distribution of all possible sample means that could be obtained from a population. Essentially, it shows us the range of values that a sample mean could take on if we were to take multiple random samples from the same population.

In your question, you are asked to prepare the sampling distribution of x-bar (sample mean) for a random sample of size n=2 drawn without replacement from a finite population of size n=5. The population in this case is the numbers 3, 5, 7, 9, and 11.

To prepare the sampling distribution, we need to consider all possible combinations of samples of size 2 that can be drawn from this population. These combinations are:

(3,5), (3,7), (3,9), (3,11), (5,7), (5,9), (5,11), (7,9), (7,11), (9,11).

Now, for each of these combinations, we calculate the sample mean (x-bar). For example, for the first combination (3,5), the sample mean is (3+5)/2 = 4. For the second combination (3,7), the sample mean is (3+7)/2 = 5, and so on.

Once we have calculated the sample mean for each combination, we can arrange them in a table and calculate the mean of these sample means. This mean is the mean of the sampling distribution, which in this case is E(x-bar) = 6.

Now, to answer the second part of your question, we need to determine if the mean of the sampling distribution (E(x-bar) = 6) is an unbiased estimator of the population mean (mu).

To determine this, we need to know the formula for an unbiased estimator. The formula is:

E(x-bar) = mu

In other words, the expected value of the sample mean is equal to the population mean.

In this case, we can see that E(x-bar) = 6, and the population mean is also 6. Therefore, E(x-bar) is an unbiased estimator of the population mean mu.

I hope this explanation helps you better understand sampling distributions